Recently I have been studying work and energy and in this chapter I encountered with the term power. In terms of work power is written as $dW/dt$. So I have a doubt that suppose power is increasing. So what does it imply in terms of energy. Does it mean that in small interval of time, energy of that body is adding up as time goes by and due to which K.E of that body is increasing? Please solve my confusion
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$\begingroup$ hello! it's good practice to choose an informative/specific title for a question. 90% of questions could go under the generic title "Please help me with this doubt". $\endgroup$– QuilloCommented Jul 29, 2023 at 11:25
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3 Answers
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Increasing power means $\frac{\text{d}^2 W}{\text{d}t^2} > 0$, which tells you nothing about the way the energy is changing in a small time interval. Same as positive acceleration not telling you whether a body is going forwards or backwards. Only if you start from an initial condition (on energy and power) and use info on $\frac{\text{d}^2 W}{\text{d}t^2}$ over a time interval from that initial point onward will you get what you are interested in by integration.
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1$\begingroup$ “Same as positive acceleration not telling you whether a body is going forwards or backwards”. Energy is a scalar quantity. Acceleration a vector quantity. Energy has no “direction”. Acceleration does. $\endgroup$– Bob DCommented Jul 29, 2023 at 12:54
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$\begingroup$ Fair enough. Add a "in a one-dimensional problem" to my comparison, if you like. $\endgroup$– kricheliCommented Jul 29, 2023 at 13:46
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So what does it imply in terms of energy.
It implies nothing more than the rate at which work is being done (rate of energy transfer by work) is increasing in time.
Does it mean that in small interval of time, energy of that body is adding up as time goes
Not sure what you mean by "energy is adding up in time". Work is energy transfer due to force times displacement. Power is the rate at which the energy is being transferred. What happens to the energy being transferred is another matter. It may dissipated as heat, stored as potential energy in a system, result in an increase in KE of the body, or any combination of these. See below.
..and due to which K.E of that body is increasing?
Maybe, maybe not.
If net work is being done on the body, then its KE will increase (re: work-energy theorem). If the net work done on the body is zero, then there would be no increase in KE of the body. An example is lifting a body of mass $m$ starting at rest on the ground and bringing it to rest at a height $h$ above the ground. You have done positive work of $mgh$, but the change in KE of the body is zero. That's because gravity did an equal amount of work of negative work $-mgh$ for a net work of zero. The work you did is stored as gravitational potential energy of the Earth-body system.
Hope this helps.
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To find the energy increase you have to integrate $dW/dt$ , so for example if power increases proportional to t so $P=k*t$ starting at $t=0$ then energy is $E(0)+k*\frac{t^2}{2}$
